Question

Calculate the number of moles of sodium oxide that should be produced when 6.50 grams of sodium reacts with oxygen gas

Answer 1

If a stoichiometric ratio is taken, then

When sodium is burned in oxygen, Na2O2 peroxide is formed, also there is an admixture of Na2O (that is, at the beginning - peroxide, and as oxygen becomes smaller, the sodium residue reduces the peroxide to oxide).

However, one mole of any of the compounds contains two moles of sodium.

$\nu(Na)=mass(Na)/Molar(Na)=> \nu(Na)=6,50g/22,99=0,2827$

v_{sodium oxide}=$\nu(Na)/2=0,14135$ moles